(a) Chứng minh rằng sin 3 x − sin x/2 cos 2 x − 1 = 2 sin x . (b) Giải phương trình cos x − sin 2 x = 0.

a) Ta có : \[VT = \frac{{\sin 3x - \sin x}}{{2{{\cos }^2}x - 1}} = \frac{{2\cos 2x.\sin x}}{{\cos 2x}} = 2\sin x = VP.\]

b) Ta có: \[\cos x - \sin 2x = 0 \Leftrightarrow \sin 2x = \sin \left( {\frac{\pi }{2} - x} \right)\]

\[ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{2} - x + k2\pi \\2x = \pi - \left( {\frac{\pi }{2} - x} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\frac{{2\pi }}{3}\\x = \frac{\pi }{2} + k2\pi \end{array} \right.(k \in \mathbb{Z}).\]